There are $2n-1$ slots/boxes in all and two objects say A and B; total number of A's are $n$ and total number of B's are $n-1$. (All A's are identical and all B's are identical.) In how many ways can we arrange A's and B's in $2n-1$ slots.
My approach: there are $2n-1$ boxes in total and for A, $n$ have to be selected, so number of ways to select $n$ A's is $C(2n-1,n)$ and can be permuted in $n!/n!$ ways, i.e., $1$. And similarly for B, $C(n-1,n-1)$ and $(n-1!)/(n-1!)$ permutations in total.
So total $$C(2n-1,n) \times 1 \times C(n-1,n-1) \times 1=C(2n-1,n).$$
Please help i am stuck.